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Proof step 30 from proof junkref-addarray-04 (previous step)

0 ≤ i, is_referencevalue(vals[i→n(i)]), i < n→i(# vals), is_array(str, r, addarray(r, ty, vals, st)), is_reftype(ty), type∃(str, ty), newref(r, st), validreflist(str, vals, ty, st), ¬ (addarray(r, ty, vals, st)[vals[i→n(i)] .refval -- _type]) .type ≤ ty ∨ junkref(str, vals[i→n(i)] .refval, addarray(r, ty, vals, st)) ⊦ vals[i→n(i)] .refval = jvmref, r = jvmref, ty .jtelemtype = abstract_type, ty .jtelemtype = void_type, vals = []

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 29:
vals[i→n(i)] = refval(r0), 0 ≤ i, is_referencevalue(vals[i→n(i)]), i < n→i(# vals), is_array(str, r, addarray(r, ty, vals, st)), is_reftype(ty), type∃(str, ty), newref(r, st), validreflist(str, vals, ty, st), ¬ (addarray(r, ty, vals, st)[r0 -- _type]) .type ≤ ty ∨ junkref(str, r0, addarray(r, ty, vals, st)) ⊦ r0 = jvmref, r = jvmref, ty .jtelemtype = abstract_type, ty .jtelemtype = void_type, vals = []

Rule argument

[speclemmaarg: ,is_referencevalue(val) → (r = val .refval ↔ val = refval(r)) ,[val, r] → [vals[i→n(i)], r0] ,T ,javavalue , ,elim-13 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_referencevalue(val) → ( r = val .refval ↔ val = refval(r))])], [EXTRATERMS : (vals[i→n(i)], r0)]]